If a < b < c < d and a, b, c, and d all are powers of 2 and all are po

Hey, so it does not matter if you have to guess the value or not. What matters is that you get a unique answer for a given statement. And you do get a unique set of numbers from statement 1 without the use of statement 2. Agreed that S2 does help, but it is not necessary to get the answer from statement 1 (or, s1 is "sufficient").

A < B < C < D

Each is a positive power of (2)

We need to know the value of D = ?

S1: A + B + C + D = 170

Since each is a power of 2, if we actually knew the values, we would be able to pull out AT LEAST one power of 2 as a common factor.

We can the divide each side of the equation by 2.

Case 1: A does NOT equal 2 to the 1st power

When we pull out the power or 2, what remains will still be a positive power of 2 from A’s value. Therefore

A/2 + B/2 + C/2 + D/2 = 85

And

Even + even + even + even = 85

But 85 is an odd numbers, a fact that would be impossible if (A/2) were even

A, the lowest value, must therefore be (2)^1

Case 2: A = (2)^1

when we take the 2 out as a common factor, what remains from the A value will be = 1

(2)^1 {1 + B/2 + C/2 + D/2} = 170

1 + B/2 + C/2 + D/2 = 85

B/2 + C/2 + D/2 = 84

*where A =(2)^1

We now have to choose three powers of 2 that sum to 84

However, if at least one of them is NOT at least (2)^6 = 64, we will not have enough value to add up to 84

{2 , 4 , 8 , 16 , 32}

Even taking the three highest that are less than 64, we get:

32 + 16 + 8 = 56

Therefore, 64 must be one of the values and must be the highest value, since any power of 2 beyond 64 will exceed the sum of 84:

B/2 + C/2 + 64 = 84

B/2 + C/2 = 20

At this point, only 16 and 4 will work as powers of 2


Therefore:

B/2 = 4 ———> B = 8
C/2 = 16 ———> C = 32
D/2 = 64 ———-> D = 128

And finally we know from above that A = 2

And we have

2 + 8 + 32 + 128 = 170

Where D = 128

This is the only possibility.

Statement 1 is sufficient alone

Statement 2 allows us to have many different values for D depending on how we apportion the powers of 2 (Exponent Rule: add the exponents when multiplying exponential terms that have the same Base)

Answer is *A*

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